Let $M$ be a smooth manifold (not necessarily orientable) and let $N=\partial M$. Is $N$ necessarily orientable?

I have no particular reason to believe that this is the case, but I wasn't able to come up with a counterexample either.

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    $\begingroup$ Take any non-orientable $K$ without boundary and set $M=K\times[0,1]$. $\endgroup$
    – user8268
    Feb 9 '12 at 22:49

The example given by azarel is a non-compact one. Here is a compact example.

We know how to make Klein bottle (i.e. we fibre a circle over a circle in a way). If we replace the fibre by disc then we get a manifold whose boundary is Klein bottle, i.e. the boundary is non-orientable.

  • $\begingroup$ Actually, I don't think this works. I don't think you can replace the fiber by a disc in a consistent manner. More specifically, since the Klein bottle is non-orientable, it's first Stiefel-Whitney class $w_1$ is nonzero, so generates $H^1(K,\mathbb{Z}/2)$. But it's known that the cube of this generator is nonzero, so the Steifel-Whitney number $\langle w_1^3,[K]\rangle$ is nonzero. It follows that the Klein bottle is not the boundary of any manifold. $\endgroup$ Apr 17 '13 at 19:53
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    $\begingroup$ I am very sorry I don't know characteristic classes (I'm an undergrad) but I think we can construct such a manifold. Lets take a solid cylinder $D^2\times [0,1]$ and identify $(x,0)$ with $(-x,1)$, where $-x$ is the reflection of $x$ about some fixed diameter of $D^2$. It is a manifold with Klein Bottle as boundary. This way we can change the fibre. Let me know if I am wrong. Thanks. $\endgroup$
    – Sunny
    Apr 17 '13 at 20:17
  • $\begingroup$ You're right! +1 from me! The mistake in my calculation is that $w_1^2 = 0$, so $w_1^3$ cannot be nonzero. Sorry about that. If you'd like, I can delete my previous comment. $\endgroup$ Apr 17 '13 at 21:00

I'm a bit rusty on this but would not something like $M\times [0,1)$ be a counterexample? (where $M$ is the Möbius band)

  • $\begingroup$ I do not get it. Boundary of boundary suppose to be empty, right? Mobius band has a non-empty boundary. $\endgroup$
    – user21965
    Dec 6 '12 at 16:33
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    $\begingroup$ There is an "open" Mobius strip (the interior of the one you're thinking of) that should work as a replacement. $\endgroup$ Apr 17 '13 at 17:55
  • $\begingroup$ I believe the boundary of $M \times [0,1)$ would (I think with slight notational abuse) be $M \times \{0,1\} \cup \partial M \times [0,1]$, which would not be orientable. It wouldn't be just a Möbius strip. $\endgroup$
    – Alex W
    Jul 5 '18 at 23:43

The general theorem is that a manifold is a boundary if and only if all of its Stiefel-Whitney numbers are zero. This can happen for both orientable and non-orientable manifolds. The Klein bottle is the simplest example of a non-orientable manifold that is a boundary.

Here is the proof that all of the Stiefel-Whitney numbers are zero for the Klein bottle.

Since the Klein bottle is a 2 dimensional surface it has only two Stiefel-Whitney numbers, the square of its first Stiefel-Whitney class and its second Stiefel Whitney class both evaluated on the mod 2 fundamental cycle.

If one thinks of a Klein bottle as a circle bundle over a circle, then the first Stiefel Whitney class is the intersection number with a fiber circle. Its self intersection number is zero. This can all be easily seen by drawing a Klein bottle as the fundamental domain of the action of the group of isometries of the plane generated by the standard lattice together with the map (x,y) -> ( x + 1/2, -y) The fiber circles are the projections of straight lines that are perpendicular to the x axis.

So the square of the first Stiefel-Whitney class is zero.

The second Stiefel Whitney class is zero because the Klein bottle has an everywhere non-zero vector field. One can also show that it is the square of the first Stiefel-Whitney class, and therefore zero,because the classifying map of the tangent bundle can be factored through the infinite real projective space.

Therefore the Klein bottle is a boundary by Thom's theorem.

But the reply above, that one can extend the fiber circle to a disk is true and this shows that the Klein bottle is a boundary directly. If you construct a Klein bottle from a cylinder by attaching the bounding circles with a reflection, then this map clearly extends to the solid cylinder.

BTW: The first Z/2 cohomology of the Klein bottle is not generated by the first Stiefel Whitney class. Its first cohomology is isomorphic to Z/2 x Z/2.

Interestingly, the square of the other cohomology class, the one which is not the first Stiefel-Whitney class is not zero. This class is the intersection number with the equatorial circle of the Klein bottle. Its self-intersection number, and therefore the square of the cohomology class, is not zero. This follows from the Intermediate Value theorem.

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    $\begingroup$ What is true is that the boundary of an of an orientable manifold is orientable. $\endgroup$
    – lavinia
    Dec 2 '13 at 13:32
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    $\begingroup$ It is tempting to think that since a boundary always has an inward pointing normal vector field that the manifold is orientable. But if the ambient manifold is not orientable then there may be no way to find an orientation of the boundary's tangent space. $\endgroup$
    – lavinia
    Dec 2 '13 at 13:36

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